【一天一道LeetCode】#68. Text Justification

一天一道LeetCode

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（一）题目

Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified.

You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ’ ’ when necessary so that each line has exactly Lcharacters.

Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.

For the last line of text, it should be left justified and no extra space is inserted between words.

For example,

words: [“This”, “is”, “an”, “example”, “of”, “text”, “justification.”]

L: 16.

Return the formatted lines as:

[

“This is an”,

“example of text”,

“justification. ”

]

Note: Each word is guaranteed not to exceed L in length.

（二）解题

题目大意：按照一定的格式对文本进行对齐。

需要注意以下几点：

1、只有一个单词直接在后面补空格，如“a”，5 输出“a ”

2、最后一组单词不需要对齐，如“a”，“b” 5 输出“a b ”

3、单词与单词之间至少要有一个空格隔开

具体 思路见代码注释：

class Solution {
public:
    vector<string> fullJustify(vector<string>& words, int maxWidth) {
        vector<string> ret;//返回值
        int width = words[0].size();
        int num = 1;
        int last = 0 ;//纪录每一次maxWidth的起始序号
        for(int i = 1 ; i<words.size() ; i++)
        {
            width+=words[i].size();
            num++;
            if(width+num-1>maxWidth)//这里要注意每个单词之间要用空格隔开
            {
                width-=words[i].size();//清除掉最后一个数
                num--;
                string temp;
                if(num==1){//只有一个单词的情况
                    temp+=words[i-1];
                    while(temp.size()!=maxWidth) temp+=" ";//在后面补齐空格
                    ret.push_back(temp);
                }
                else//多个单词，但不是结尾的情况
                {
                    int blankWidth = maxWidth - width;
                    int gap = 0;
                    for(int j = last ; j < i ; j++)
                    {
                        if(j==i-1) gap=0;//最后一个单词后面不加空格
                        else {
                            gap = blankWidth/(num-1);//每一次进来都要算需要增加多少空格
                            gap = blankWidth%(num-1)>0?gap+1:gap;//保证均匀分布
                            blankWidth -=gap;
                        }
                        temp+=words[j];
                        while(gap>0&&gap--) temp+=" ";
                        num--;
                    }
                    ret.push_back(temp);
                }
                //初始化下一个循环
                last=i;
                width = words[i].size();
                num=1;
            }
        }
        if(last<words.size()){//考虑末尾不足maxWidth的情况
            int j = last;
            string temp;
            while(j<words.size()){//先按一个空格添加words
                temp+=words[j++];
                if(temp.size()<maxWidth) temp+=" ";
            }
            while(temp.size()<maxWidth) temp+=" ";//最后不足maxWidth就用空格补足
            ret.push_back(temp);
        }
        return ret;
    }
};